matrix multiplication
Peter Otten
__peter__ at web.de
Tue Feb 27 06:08:19 EST 2018
Seb wrote:
> On Tue, 27 Feb 2018 12:25:30 +1300,
> Gregory Ewing <greg.ewing at canterbury.ac.nz> wrote:
>
>> Seb wrote:
>>> I was wondering is whether there's a faster way of multiplying each
>>> row (1x3) of a matrix by another matrix (3x3), compared to looping
>>> through the matrix row by row as shown in the code.
>
>> Just multiply the two matrices together.
>
>> If A is an nx3 matrix and B is a 3x3 matrix, then C = A @ B is an nx3
>> matrix where C[i] = A[i] @ B.
>
>> (This is a property of matrix multiplication in general, nothing
>> special about numpy.)
>
> I think that's only true if B is the same for every row in A. In the
> code I posted, B varies by row of A.
Yeah, you would have to substitute the N 3x3 matrices with an Nx3x3 tensor,
though I don't know if numpy provides an op such that
Nx3 op Nx3x3 --> desired result
or
op(Nx3, Nx3x3) --> desired result
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